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  • IEEE P1918.1.12019-05-01 20:42:14

    IEEE P1918.1.1Haptic Codecs for the Tactile Internet Task GroupProposal for Tactile CodecTUM Vibrotactile Perceptual Codec based on DWT and SPIHT (TUM-VPC-DS)DCN: HC_NGS_19-1-r0_Proposal_for_Tactile_CodecDate: 2019-3-29AbstractThis document describes a pr

  • LeetCode算法题-1-bit and 2-bit Characters(Java实现)2019-04-10 08:51:49

    这是悦乐书的第302次更新,第321篇原创 01 看题和准备 今天介绍的是LeetCode算法题中Easy级别的第170题(顺位题号是717)。有两个特殊字符,第一个字符可以用一个比特0表示,第二个字符可以用两个比特(10或11)表示。现在给出一个由比特位组成的数组,判断其最后一个字符是否是一位字符。数组的

  • ACM-ZOJ2729-Sum Up2019-04-07 17:45:19

    提交网址:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=2729 题目:   Vivid has stored a piece of private information, which consisted of a serial of integers in a secret number format. All the stored numbers are in the range [-63, 63]. So every n

  • Exponentiation POJ-10012019-04-06 13:40:31

    http://poj.org/problem?id=1001 1 //10000000 100000 2 3 #include<iostream> 4 #include<cstring> 5 using namespace std; 6 7 char ts[7],tts[7]; //用于处理输入,删去点,签到0,点后0,统计小数点后数的个数保存至origu 8 short orig[7]

  • CS 2505 Data Lab: Data and Bitwise Operations2019-04-05 19:44:10

    CS 2505 Spring 2019Data Lab: Data and Bitwise OperationsAssigned: March 19Due: Tuesday March 29, 23:59Ends: Friday April 5, 23:591 IntroductionThe purpose of this assignment is to become more familiar with bit-level representations of integers. You’lldo t

  • 717. 1-bit and 2-bit Characters--easy2019-04-04 13:47:52

    We have two special characters. The first character can be represented by one bit 0. The second character can be represented by two bits (10 or 11). Now given a string represented by several bits. Return whether the last character must be a one-bit charac

  • CPCI高速信号处理板CPCI6U-4DSP-C64552019-04-01 18:50:38

    CPCI-4DSP-C6455是一款6U CPCI标准通用信号处理板卡,核心处理器件为4片TI公司的DSP芯片TMS320C6455和1片Xilinx公司的FPGA芯片XC4VSX55,每个DSP可外挂512MB DDR2 SDRAM。该板卡对外接口包括1个32/64bits 33/66MHz PCI 接口、2个PMC子卡接口、4个4×1.25Gbps SRIO(Serial RapidIO)接

  • LeetCode - 191. Number of 1 Bits2019-03-31 16:50:02

    Write a function that takes an unsigned integer and return the number of '1' bits it has (also known as the Hamming weight). Example 1: Input: 00000000000000000000000000001011Output: 3Explanation: The input binary string 000000000000000000000000

  • 关于位操作符的小练习2019-03-31 13:47:35

    写一个函数返回参数二进制中 1 的个数。 比如: 15 0000 1111 4 个 1 程序原型: int count_one_bits(unsigned int value) { // 返回 1的位数 } 这里主要利用位操作符的作用 #include <stdio.h> #include <stdlib.h> int count_one_bits(unsigned int num) { int i = 0; int

  • char是所有类型中最短的 char多为8位,2019-03-15 11:48:21

      https://en.wikipedia.org/wiki/C_data_typesIn practice, char is usually eight bits in size and short is usually 16 bits in size (as are their unsigned counterparts).char多为8位,As char's size is always the minimum supported data type, no other data typ

  • [Swift]LeetCode717. 1比特与2比特字符 | 1-bit and 2-bit Characters2019-03-11 12:54:50

    We have two special characters. The first character can be represented by one bit 0. The second character can be represented by two bits (10 or 11). Now given a string represented by several bits. Return whether the last character must be a one-bit charac

  • [Swift]LeetCode693. 交替位二进制数 | Binary Number with Alternating Bits2019-03-09 18:47:51

    Given a positive integer, check whether it has alternating bits: namely, if two adjacent bits will always have different values. Example 1: Input: 5Output: TrueExplanation:The binary representation of 5 is: 101  Example 2: Input: 7Output: FalseExplanation

  • LeetCode 693 Binary Number with Alternating Bits 解题报告2019-03-09 09:54:18

    题目要求 Given a positive integer, check whether it has alternating bits: namely, if two adjacent bits will always have different values. 题目分析及思路 给定一个正整数,判断它的二进制形式是否是01交替出现。如果是则返回True,否则返回False。可以先获得该数的二进制形式

  • LeetCode 762 Prime Number of Set Bits in Binary Representation 解题报告2019-03-08 10:40:44

    题目要求 Given two integers L and R, find the count of numbers in the range [L, R] (inclusive) having a prime number of set bits in their binary representation. (Recall that the number of set bits an integer has is the number of 1s present when written i

  • java多线程---synchronizedsynchronized分析2019-03-03 15:51:54

    synchronized分析 使用方法 1、修饰方法 public synchronized void test(){ } 2、修饰对象(锁定的为什么一定高要是对象?) public void test2(){ //Todo synchronized (this){ } //todo } //由于是修饰对象,syn的范围由所修饰对象的生命周期来 锁

  • DES加密解密工具2.1及其代码——支持字符串及文件加密,支持3重DES2019-03-02 11:53:31

    转自:http://www.cnblogs.com/erwin/archive/2009/04/14/1435288.html DES加密解密工具2.1及其代码——支持字符串及文件加密,支持3重DES 作者:尹学渊 版权所有,转载请务必注明出处 一.DES 2.1 说明 最近非常的忙,事情都已经堆着了。所以迟迟没有更新博客,请各位需要代码的朋友谅解

  • Hamming Distance2019-02-26 22:50:16

    Hamming Distance The Hamming distance between two integers is the number of positions at which the corresponding bits are different. Given two integers x and y, calculate the Hamming distance. Example Input: x = 1, y = 4 Output: 2 Explanation: 1 (0 0

  • 【Codeforces 484A】Bits2019-02-25 20:37:29

    【链接】 我是链接,点我呀:) 【题意】 让你求出l~r当中二进制表示1的个数最多的数x 【题解】 最多有64位 我们可以从l开始一直增大到r 怎么增大? 找到l的二进制表示当中0所在的位置 假设i这一位的0经过加法变成了1 那么我们再从低位到高位依次枚举那一位为1就好 然后把这个

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